Isomorphism between $H^1(\mathbb{P}^n,\mathcal{O}^*)$ and $H^2(\mathbb{P}^n,\mathbb{Z})$

Question

In Griffiths and Harris "Principles of Algebraic Geometry" in page 145 they write

$$Pic(\mathbb{P}^n)\simeq H^2(\mathbb{P}^n,\mathbb{Z})=\mathbb{Z}$$

and they justify it by saying $H^1(\mathbb{P}^n,\mathcal{O})=0$.

I have two questions:

• They are sending to a nonexistent reference such as Chapter 1 section 7. Where is the real one?

• Second, for the isomorphism to be true we need also the second cohomology group to vanish, that is $H^2(\mathbb{P}^n,\mathcal{O})=0$ as well. Is that correct? If yes where could I find a reference for that?

Thanks is advance.

Answer 1

I don't know if it worth as an answer but...

Both questions are solved with the so called Bott-formulae A particular case can be found at the top of page 108 (section 7 of Chap 0) of Griffiths & Harris. This is as follows:

Bott Formulae: $$ \dim {\rm H}^q(\mathbb{P}^n, \Omega^p_{\mathbb{P}^n}\otimes \mathcal{O}_{\mathbb{P}^n}(k)) = \begin{cases} \binom{k+n-p}{k}\binom{k-1}{p} & {\rm for }\quad q=0,\, 0\leq p \leq n,\, k>p\\ 1 & {\rm for }\quad k=0,\, 0\leq p =q\leq n\\ \binom{p-k}{-k}\binom{-k-1}{n-p} & {\rm for }\quad q=n,\, 0\leq p \leq n,\, k<p-n\\ 0 & {\rm otherwise }\end{cases} $$

This result can be found in Okonek, Schneide, Splinder - Vector Bundles on Complex Projective Spaces page 8.

The result in Griffiths & Harris is the particular case where $k=0$.

Answer 2

You can show the relevant cohomology groups vanish using Hodge theory. Letting $h^{p,q}=\dim_{\mathbb C}H^q(\mathbb P^n,\Omega^p)$ and $b^i=\dim_{\mathbb C}H^i(\mathbb P^n,\mathbb C)$, we get the following (among other things) \begin{align*} b^1 = h^{0,1}+h^{1,0} && b^2=h^{0,2}+h^{1,1}+h^{2,0} && h^{0,2}=h^{2,0} \end{align*} At the same time, we know that $b^1=0$, so $h^{0,1}=h^{1,0}=0$, and that $b^2=1$. Since $h^{0,2}=h^{2,0}$, this means that $h^{1,1}=1$ and $h^{0,2}=0=h^{2,0}$. In particular, we've shown that \begin{align*} 0=h^{0,1}=\dim H^1(\mathbb P^n, \mathscr O) && 0=h^{0,2}=\dim H^1(\mathbb P^n,\mathscr O) \end{align*} so we get the desired isomorphism $H^1(\mathbb P^n,\mathscr O^*)\to H^2(\mathbb P^n,\mathbb Z)$.